Question

# Part A The activation energy of a certain reaction is 43.9 kJ/mol . At 28  ∘C ,...

Part A

The activation energy of a certain reaction is 43.9 kJ/mol . At 28  ∘C , the rate constant is 0.0130s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Part B

Given that the initial rate constant is 0.0130s−1 at an initial temperature of 28  ∘C , what would the rate constant be at a temperature of 100.  ∘C for the same reaction described in Part A?

we have to use the equation

ln(k2/k1) = (Ea/R) · (1/T1 - 1/T2)

K2/K1 = 2

Ea = activation energy give =43.9 kJ/mol = 43900 J/mol

R =gas constant 8.314 J/mol-K

T1 = 273 + 28 = 301 K

put all these in above equation

ln (2) = 43900/8.314 [1/301 -1/T2]

0.693 = 5280.25 [0.0033 -1/T2]

[0.0033 -1/T2] = 0.693 / 5280.25

[0.0033 -1/T2] = 0.00013

1/T2 = 0.00317

T2 = 1/0.00317

T2 = 315.46 K

partB

use the same equation we need to find out K2

ln(k2/k1) = (Ea/R) · (1/T1 - 1/T2)

ln [K2 / 0.0130] = 43900 / 8.314 [1/301 - 1/373 ]

ln [K2 / 0.0130] = 5280.25 [ 0.0033 - 0.0026]

ln [K2 / 0.0130] = 3.7

[K2 / 0.013] = e3.7

[K2 / 0.013] = 40.4473

K2 = 0.5258 s-1